I apologize. The above answer is for a z-score of 4. The calculation was too close to 1 as to have problems with rounding errors!

The true answer is as follows:

`p(h<=80) = 1-2.6904*10^-11`

In other words, the likelihood that a woman is less than 80" tall is effectively 1.

However, this probability tells us that the population mean and standard deviation are certainly not true. Based on this number, a woman would never be tall enough to need to lean to get past a door; however, I have certainly seen it before!

Let's start by filling a gap in the problem. The standard door in America has the following dimensions: 36" x 80" (width x height). In other words, in order for a woman to get through a door without trouble, she would need to be maximally 80" tall.

Now, let's move on to the meat of the problem: how can we use the normal distribution to tell us the percentage of women that fit through a door? We are given that the following information:

`mu = 63.6`

`sigma = 2.5`

Here, `mu` is the mean for women's height, and `sigma` is the standard deviation. Using this information, we can take a given height, calculate its "z-score" (how many standard deviations above or below the mean a given value is), and determine the percentage of people fall below that height.

Thus, what we want to do is calculate the z-score for a height of 80" and use that to determine how many women fall below that height. To calculate the z-score, we use the following equation:

`z = (h-mu)/sigma`

Here, `h` is the height that we are testing, which is 80". We can now substitute numbers in for each variable in this equation, allowing us to calculate `z`:

`z = (80 - 63.6)/2.5 = 16.4/2.5 = 6.56`

Thus, we are 6.56 standard deviations above the mean! This z-score is quite high, which we can see by the fact that most z-score tables only go up to 4. Many calculators online simply say the probability of being below this z-score is 1.

At this point, you must use a calculator or computer with a high level of precision to find the probability that a woman's height will be below the z-score of 6.56. Using Matlab, I was able to calculate that the probability of being below the z-score is 0.999968.

In other words, the probability that a woman can fit through a standard door is 0.999968.